## Abstract

A novel, simple tuning mechanism for single-mode, pseudo-external-cavity diode lasers has been developed. The model calculations predict that the laser can be tuned continuously by as much as 300 GHz in the vicinity of the chosen frequency without locking it to an external cavity. Experimentally, the continuous tuning range is approximately 120 GHz at constant current and temperature for the 7-cm-long pseudo-external cavity; this is several times more than previously reported. A turning wedge inside the laser cavity is used as the tuning element. The laser is based on a commercial laser diode chip, and a diffraction grating is used for feedback. The total tuning range depends on the laser diode type and can be up to 20 nm.

© 1995 Optical Society of America

Full Article |

PDF Article
### Equations (18)

Equations on this page are rendered with MathJax. Learn more.

(1)
$$\delta \phantom{\rule{0em}{0ex}}\lambda \phantom{\rule{0em}{0ex}}/\phantom{\rule{0em}{0ex}}\Lambda =\delta \phantom{\rule{0em}{0ex}}\lambda \frac{2\phantom{\rule{0.1em}{0ex}}({L}_{0}+\Delta L)}{{({\lambda}_{\text{CAV}})}^{2}}\phantom{\rule{0.2em}{0ex}}.$$
(2)
$$\Delta L=L\phantom{\rule{0.1em}{0ex}}(\varphi )-L\phantom{\rule{0.1em}{0ex}}({\varphi}_{0})\phantom{\rule{0.2em}{0ex}},$$
(3)
$${\lambda}_{\text{CAV}}={\lambda}_{0}\phantom{\rule{0.1em}{0ex}}(1+\frac{\Delta L}{{L}_{0}})\phantom{\rule{0.2em}{0ex}},$$
(4)
$${\lambda}_{\text{DG}}=\frac{2sin\u03f4}{N\times {10}^{3}}\phantom{\rule{0.2em}{0ex}},$$
(5)
$$\delta \phantom{\rule{0em}{0ex}}\lambda ={\lambda}_{\text{DG}}-{\lambda}_{\text{CAV}}\phantom{\rule{0.2em}{0ex}},$$
(6)
$$L=n{L}_{1}+{L}_{2}\phantom{\rule{0.2em}{0ex}},$$
(7)
$${L}_{1}=h\phantom{\rule{0.1em}{0ex}}[sin\psi -cos\psi tan\phantom{\rule{0em}{0ex}}(\psi -\alpha )\phantom{\rule{0.1em}{0ex}}]\phantom{\rule{0.2em}{0ex}},$$
(8)
$${L}_{2}=(X-{L}_{1})\phantom{\rule{0.1em}{0ex}}[cos({\varphi}_{1}-{\psi}_{1})-sin({\varphi}_{1}-{\psi}_{1})tan\u03f4]\phantom{\rule{0.2em}{0ex}},$$
(9)
$$X=D\phantom{\rule{0.1em}{0ex}}[cos(\varphi -\psi )+sin(\varphi -\psi )\times tan({\u03f4}_{0}-\psi +\varphi -{\phi}_{0})\phantom{\rule{0.1em}{0ex}}]\phantom{\rule{0.2em}{0ex}},$$
(10)
$$h=\frac{d}{2\phantom{\rule{0.2em}{0ex}}sin(\alpha /2)}\phantom{\rule{0.2em}{0ex}},$$
(11)
$${\u03f4}_{0}=arcsin(\frac{{\lambda}_{0}}{2}N\times {10}^{3})\phantom{\rule{0.2em}{0ex}},$$
(12)
$${\psi}_{1}=\psi -\alpha \phantom{\rule{0.2em}{0ex}},$$
(13)
$$sin\varphi =nsin\psi \phantom{\rule{0.2em}{0ex}},$$
(14)
$$sin{\varphi}_{1}=nsin{\psi}_{1}\phantom{\rule{0.2em}{0ex}},$$
(15)
$$\phi =\varphi -{\varphi}_{1}-\alpha \phantom{\rule{0.2em}{0ex}},$$
(16)
$${\phi}_{0}={\varphi}_{0}-{\varphi}_{1}({\varphi}_{0})-\alpha \phantom{\rule{0.2em}{0ex}},$$
(17)
$$\u03f4={\u03f4}_{0}-{\phi}_{0}+\varphi -{\varphi}_{1}-\alpha \phantom{\rule{0.2em}{0ex}}.$$
(18)
$$\Delta \phantom{\rule{0em}{0ex}}\lambda ={({\lambda}_{\text{CAV}})}_{max}-{({\lambda}_{\text{CAV}})}_{min}\phantom{\rule{0.2em}{0ex}},$$